Question

Let $f,g:[ - 1,2] \to R$ be continuous functions which are twice differentiable on the interval $( - 1,2)$. Let the values of f and g at the points $- 1,0$ and $2$ be as given in the following table:

3;x = - 1 \\\ 6;x = 0 \\\ 0;x = 2 \\\ \right.$$and $$g(x) = \left\\{ 0;x = - 1 \\\ 1;x = 0 \\\ 0;x = - 1 \\\ \right.$$ In each of the intervals $$( - 1,0)$$ and $$(0,2)$$, the function $$(f - 3g)''$$never vanishes. Then the correct statement(s) is (are) THIS QUESTION HAS MULTIPLE CORRECT OPTIONS A.$$f'(x) - 3g'(x) = 0$$ has exactly three solutions in $$( - 1,0) \cup (0,2)$$ B.$$f'(x) - 3g'(x) = 0$$ has exactly one solution in $$( - 1,0)$$ C.$$f'(x) - 3g'(x) = 0$$ has exactly one solution in$$(0,2)$$ D.$$f'(x) - 3g'(x) = 0$$ has exactly two solutions in $$( - 1,0)$$ and exactly two solutions in $$(0,2)$$

Answer 1

First form the equation as per the given information in the question. And then apply Rolle’s theorem to calculate their differentiation, as Rolle's theorem states that if a function f is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ such that $f(a) = f(b)$, then $f'(x) = 0$ for some x with $a \leqslant x \leqslant b$. Hence, after differentiation of the above formed term we can reach the solution required.

Complete step-by-step answer:
As the given conditions are of $f,g:[ - 1,2] \to R$ be continuous functions which are twice differentiable on the interval $( - 1,2)$.
And the values of f and g at the points $- 1,0$ and $2$ be as given in the following table:

3;x = - 1 \\\ 6;x = 0 \\\ 0;x = 2 \\\ \right.$$and $$g(x) = \left\\{ 0;x = - 1 \\\ 1;x = 0 \\\ 0;x = - 1 \\\ \right.$$ Now, let the new equation be $$h(x) = f(x) - 3g(x)$$ Now calculate the value of $$h(x)$$at various points $$ - 1,0$$ and $$2$$

h( - 1) = f( - 1) - 3g( - 1) \\
= 3 - 0 \\
= 3 \\

$$Similarly calculate for remaining point as,$$

h(0) = f(0) - 3g(0) \\
= 6 - 3(1) \\
= 3 \\

$$And,$$

h(2) = f(2) - 3g(2) \\
= 0 - 3( - 1) \\
= 3 \\

Hence, now calculate the differentiation of $$h'(x)$$, so it can be given as Let us calculate the value of $$h'(x)$$applying Rolle’s for interval $$( - 1,0)$$. As the function satisfies all the Rolle’s condition so it can be given as, $$h'(x) = 0$$ Thus, has exactly one root in the interval $$( - 1,0)$$ So, $$f'(x) - 3g'(x) = 0$$ has exactly one solution in $$( - 1,0)$$ Let us calculate the value of $$h'(x)$$applying Rolle’s for interval$$(0,2)$$. As the function satisfies all the Rolle’s condition so it can be given as, $$h'(x) = 0$$ Thus, has exactly one root in the interval $$(0,2)$$ So, $$f'(x) - 3g'(x) = 0$$ has exactly one solution in $$(0,2)$$ **Hence, option (B) and option (C) require the correct answer.** **Note:** Rolle's theorem, in analysis, special case of the mean-value theorem of differential calculus. Rolle's theorem states that if a function f is continuous on the closed interval $$[a,b]$$ and differentiable on the open interval $$(a,b)$$ such that $$f(a) = f(b)$$, then $$f'(x) = 0$$ for some x with $$a \leqslant x \leqslant b$$. Calculate the function $$h\left( x \right)$$properly and make use of it’s derivative without any mistake, also use the value of given functions at stated points properly.